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  • Need solution for RD Sharma maths class 12 chapter Differential Equation exercise 21.3 question 15

Need solution for RD Sharma maths class 12 chapter Differential Equation exercise 21.3 question 15

Answers (1)

Answer:

y^{2}=4 a(x+a)  is a solution of differential equation

Hint:

Differentiate the solution and modify the given differential equation

Given:

y^{2}=4 a(x+a) 

Solution:

Differentiating on both sides with respect to x

\begin{aligned} &\frac{d}{d x}\left(y^{2}\right)=\frac{d}{d x}(4 a(x+a)) \\\\ &2 y \frac{d y}{d x}=4 a \end{aligned}

\begin{aligned} &\frac{d y}{d x}=\frac{4 a}{2 y} \\\\ &\frac{d y}{d x}=\frac{2 a}{y} \end{aligned}                                ...............(i)

Put value of equation (i) in given problem as follows

\begin{aligned} &y\left[1-\left(\frac{d y}{d x}\right)^{2}\right]=2 x \frac{d y}{d x} \\\\ &L H S=y\left[1-\left(\frac{d y}{d x}\right)^{2}\right] \end{aligned}

\begin{aligned} &=y\left[1-\left(\frac{2 a}{y}\right)^{2}\right] \\\\ &=y-\frac{4 a^{2} y}{y^{2}} \\\\ &=y-\frac{4 a^{2}}{y} \end{aligned}                        ..............(ii)

\begin{aligned} &R H S=2 x \frac{d y}{d x} \\\\ &=2 x\left(\frac{2 a}{y}\right) \\\\ &=\frac{4 a x}{y} \end{aligned}                                     ..............(iii)

Now,  y^{2}=4 a(x+a)

\begin{aligned} &y^{2}=4 a x+4 a^{2} \\\\ &\frac{y^{2}-4 a^{2}}{4 a}=x \end{aligned}                                    ...........(iv)

Put value of equation (iv) in equation (iii)

\begin{aligned} &=\frac{4 a}{y}\left(\frac{y^{2}-4 a^{2}}{4 a}\right) \\\\ &=\frac{y^{2}}{y}-\frac{4 a^{2}}{y} \\\\ &=y-\frac{4 a^{2}}{y} \end{aligned}                                ............(v)

From (ii) and (v)

L H S=R H S

Thus, y^{2}=4 a(x+a) is a solution of differential equation.

Posted by

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